Question
Find the maximum value of the function f(x) = xΒ³ β
3xΒ² β 9x + 20.Solution
We are given the function: f(x) = xΒ³ β 3xΒ² β 9x + 20 We need to find its maximum value. Differentiate f(x): fβ²(x) = 3xΒ² β 6x β 9 Set fβ²(x) = 0: 3xΒ² β 6x β 9 = 0 Divide all terms by 3: xΒ² β 2x β 3 = 0 Factor: (x β 3)(x + 1) = 0 β x = 3, β1 Differentiate again: fβ³(x) = 6x β 6 Now test the sign of fβ³(x) at critical points:
- At x = β1:
fβ³(β1) = 6(β1) β 6 = β12 β Negative β local maximum - At x = 3:
fβ³(3) = 6(3) β 6 = 12 β Positive β local minimum
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